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STALIN [3.7K]
3 years ago
9

Explain Newton’s 3 laws of motion by using the example of a rollercoaster?

Physics
1 answer:
ivolga24 [154]3 years ago
5 0

Newton's First law states the Law of Inertia. The Law of Inertia says that an object at rest must stay at rest unless an applied force is acted upon it. Roller coasters run by this law, they must be pulled by a motor to get started. ... At ride will continue until another opposite force is applied to the ride, the breaks.

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A train accelerates from 23m/s to 190m/s in 54 seconds. What was its acceleration?
Ierofanga [76]

Answer:

Use the method on the image and solve it.

3 0
3 years ago
1. ¿Qué presión se ejerce sobre cada una de las cuatro patas de una mesa si su masa es de 20 kg y
mestny [16]
Tenemos.

Masa de la mesa = 20kg
Masa encima = 10kg
Masa total = 30kg
Area de cada pata = 20cm² = 20cm² * 1/10000cm² * 1m² =0,002m²

Presió(P)n que se ejerce sobre cada pata.

P =F/A
P = Masa por gravedad/A Masa = 30kg Gravedad =9,8m/s²
P =(30kg * 9,8m/s²)/0,002m²
P = 294kg * m/s²/0,002m² Pero kg *m/s² = Nw
P = 294Nw0,002m²
P = 147000 Nw/m² Pero Nw/m² = pa
p =1,47 * 10⁵ pa

Respuesta.
La presión que se ejerce sobre cada pata es de 1,47 * 10⁵pa
8 0
2 years ago
As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil
galina1969 [7]

Answer:

Second option 6.3 N at 162° counterclockwise from  

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

-F_3sin(b) + F_1 = 0

For the address and we have:

-F_3cos(b) + F_2 = 0

The forces F_1 and F_2 are known

F_1 = 5.7\ N\\\\F_2 = 1.9\ N

We have 2 unknowns (F_3 and b) and we have 2 equations.

Now we clear F_3 from the second equation and introduce it into the first equation.

F_3 = \frac{F_2}{cos (b)}

Then

-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°

Then we find the value of F_3

F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N

Finally the answer is 6.3 N at 162° counterclockwise from  

F1->

7 0
3 years ago
Why are simple everyday actions considered
Oxana [17]
They create energy ……………….
8 0
3 years ago
Read 2 more answers
Suppose a 48-N sled is resting on packed snow. The coefficient of kinetic friction is 0.10. If a person weighing 660 N sits on t
Annette [7]

Assume the snow is uniform, and horizontal.

Given:

coefficient of kinetic friction = 0.10 = muK

weight of sled = 48 N

weight of rider = 660 N

normal force on of sled with rider = 48+660 N = 708 N = N

Force required to maintain a uniform speed

= coefficient of kinetic friction * normal force

= muK * N

= 0.10 * 708 N

=70.8 N


Note: it takes more than 70.8 N to start the sled in motion, because static friction is in general greater than kinetic friction.


8 0
3 years ago
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